# Free monoids and full transformation monoids

I asked this question on math stack exchange, but I didn't get any responses. So, now I am motivated to ask it here. Is the class of free monoids first order axiomatizable? And what about the class of full transformation monoids, in other words monoids that are isomorphic to a monoids of all functions over a set S to itself under composition?

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For the second question, the answer is clearly "no": no transformation monoid is countably infinite, and so the Lowenheim-Skolem theorem kills this one. I'm sure the answer to the first question is "no" as well, but I don't see the proof yet. – Noah Schweber May 29 '14 at 3:24
You might try axiomatizing lack of relations for free monoids. Cancellation properties are a good start folllowed by ab=cd iff one or more of them is a unit or else there are r s and t with a and b and c and d being some finite combinations of r, s, and t. – The Masked Avenger May 29 '14 at 6:57
The answer to the first question is “no” by the same argument as in mathoverflow.net/a/131986/12705. – Emil Jeřábek May 29 '14 at 9:56
Related question mathoverflow.net/questions/17483/… – Benjamin Steinberg May 29 '14 at 18:34

Let F be a non-principal ultrafilter on the natural numbers. Let M be the free monoid on the one element set {1}. Let $\mu$ be the element of the ultrapower $M^{F}$ determined by $\mu(n):=1\ldots 1$ where $n$-many $1$'s occur here. Essentially, $\mu$ is an infinite word. It is then easy to see that $\mu$ has infinitely many prefixes and therefore $M^{F}$ cannot be a free monoid. For example, we could take $\nu(n):=1$ for all $n$ and $\xi(n)$ to be the empty word $()$ when $n=0$ and $\mu(n-1)$ otherwise. Then $\mu=\nu\xi$ in $M^{F}$ (the set on which they disagree contains just the number $0$). We can now perform a shift of this example in any number of ways to obtain infinitely many prefixes of $\mu$.